Question

Find the roots of the quadratic equation 3x 2 - 2√6x+2=0

Answer 1

3x^2-√6x-√6x+2=0 √3x(√3x-√2)-√2(√3x-√2)=0 (√3x-√2)(√3x-√2)=0 x=√2/√3 and x=√2/√3

Answer 2

Answer:

Roots of the given equation are $\frac{2\sqrt{6} }{4}$   and  $\frac{2\sqrt{6} }{4}$

Step-by-step explanation:

Given equation   $3x^{2} -2\sqrt{6} x +2 =0$

compare the given equation with $ax^{2} + bx + c =0$  

⇒ $a =3$, $b = -2\sqrt{6}$  and $c = 2$  

⇒ $b^{2} -4ac =$ $(-2\sqrt{6} ) ^{2} - 4 ( 3) ( 2)$  

                 $= 4(6) -24$

                 =  24 -24 = 0  

⇒ roots of given equations are equal and real      

x = [- b ±  $\sqrt{b^{2} - 4ac }$  ] / 2a  

   = [ - ( -2$\sqrt{6}$) ± $\sqrt{(-2\sqrt{6} )^{2} - 4(3)(2) }$ / 2(3)

   =   4($\sqrt{6}$) ± $\sqrt{0}$ / 6  

   =  4 $\sqrt{6}$ / 6

   = $\frac{2\sqrt{6} }{3}$

⇒ roots of the given equation are $\frac{2\sqrt{6} }{3 }$  and $\frac{2\sqrt{6} }{3}$